Thinking Like an Engineer: An Active Learning Approach (4th Edition)
4th Edition
ISBN: 9780134639673
Author: Elizabeth A. Stephan, David R. Bowman, William J. Park, Benjamin L. Sill, Matthew W. Ohland
Publisher: PEARSON
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Textbook Question
Chapter 12, Problem 1RQ
- 1. For a simple capacitor with two f lat plates, the capacitance (C) [F] can be calculated by
C =εrε0Ad
Where
εo = 8.854 x 10-12 [F /m] (the permittivity of free space in farads per meter)
εr = relative static permittivity, a property of the insulator [dimensionless]
A = area of overlap of the plates [m2]
d = distance between the plates [m]
Several experimental capacitors were fabricated with different plate areas (A), but with the same interpolate distance (d = 1.2 mm) and the same insulating material, and thus the same relative static permittivity (εr). The capacitance of each device was measured and plotted versus the plate area. The graph and trendline follow. The numeric scales were deliberately omitted.
- a. What are the units of the slope (19.15)?
- b. If the capacitance is 3 nanofarads [nF], what is the area (A) of the plates?
- c. What is the relative static permittivity of the insulating layer?
- d. If the distance between the plates were doubled, how would the capacitance be affected?
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Chapter 12 Solutions
Thinking Like an Engineer: An Active Learning Approach (4th Edition)
Ch. 12.3 - Fluid A as a dynamic viscosity of 0.5 centipoise...Ch. 12.3 - Fluid A has a dynamic viscosity of 0.5 centipoise...Ch. 12.3 - Fluid A has a dynamic viscosity of 0.5 centipoise...Ch. 12.4 - You have three springs, with stiffness 1,2 and 3...Ch. 12.4 - You have three resistors with resistance 2,2, and...Ch. 12.4 - You have four 60-nanofarad [nF] capacitors. Using...Ch. 12.4 - You have three 120 millihenry [mH] inductors. Can...Ch. 12.5 - The graph shows the ideal gas law relationship...Ch. 12.5 - The preceding graph shows the ideal gas Jaw...Ch. 12.6 - The decay of a radioactive isotope was tracked...
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